Theorem 5.2.2. Assume that a in Theorem 5.2.1 is horizontally periodic. Then v has an extension to a global, strong solution to the primitive equations (5.2.1)ontoL×(0,∞) that is even real analytic.
Remark 5.2.3. Note that since the perturbation term a2 fails to be continuous with respect to the horizontal variables, it holds that the mapping Sa2 :t7→S(t)a2 satisfies
Sa2 ∈Cb((0,∞);Xσ∞,1(L)),
but fails to be continuous att = 0. However, it holds thatv−Sa2 ∈Cb([0, T);Xσ∞,1(L)).
Since the hydrostatic Stokes semigroup admits L1-Lp-smoothing for the vertical regu-larity, see Theorem 4.2.1, we also show additional regularity for the local solution under the assumption that the initial data belongs to L∞HLpz(L)2 for some p ∈ (1,∞]. This additional regularity also plays an important role in obtaining a global extension.
Lemma 5.2.4. Let p∈(1,∞] and assume that a in Theorem 5.2.1 further satisfies (i) a ∈L∞HLpz(L)2, (ii) a∈Xσ∞,p(L), or (iii) a∈BU C(R2×[−h,0])2. Then the solution v has the additional regularity
(i) t(1−1/p)/2v, t1−1/(2p)∇v ∈L∞(0, T;L∞HLpz(L))2,
(ii) t(1−1/p)/2v ∈C([0, T);Xσ∞,p(L))and t1−1/(2p)∇v ∈L∞(0, T;L∞HLpz(L))2, (iii) t1/2v ∈C([0, T);BU C(R2×[−h,0]))2 and t∇v ∈L∞(0, T;L∞(L))2 for all 0< T <∞.
Observe that, since u= (v, w) satisfies divu = 0, the nonlinear term of the primitive equations can be written as
(v· ∇H)v+w∂zv = (u· ∇)v =∇ ·(u⊗v).
The key components to our proof of the existence of local mild solutions are the following estimates for the hydrostatic Stokes semigroup applied to this nonlinear term.
Lemma 5.2.5. Let α ∈[0,1). Then there exists a constant C = Cα > 0 such that for all v1, v2 ∈L∞,1σ (L) with ∇v1,∇v2 ∈L∞HL1z(L)2 it holds that
kS(t)P∇ ·(u1⊗v2)k∞,1 ≤Ct−(1−α)/2Gα(v1, v2), where
Gα(v1, v2) := (kv1k1,∞,1kv2k∞,1+kv2k1,∞,1kv1k∞,1)1−α(kv1k1,∞,1kv2k1,∞,1)α and ui = (vi, wi) with wi =wi(vi) as in (5.0.2) for i∈ {1,2}.
5.2 The case of Neumann boundary conditions Proof. Using the notation k·k∞,∞ := k·kL∞(L), we apply the Poincar´e inequality for average-free functions in the vertical direction to obtain, for i∈ {1,2}, that
kvi−vik∞,∞≤Ck∂zvik∞,1, kvik∞,∞ ≤ kvik∞,1, which implies
kvik∞,∞≤ kvi−vik∞,∞+kvik∞,∞ ≤Ckvik1,∞,1. The anisotropic H¨older inequality then yields
k∇(v1 ⊗v2)k∞,1 ≤ k∇v1k∞,1kv2k∞,∞+kv1k∞,∞k∇v2k∞,1
≤Ckv1k1,∞,1kv2k1,∞,1,
kv1⊗v2k∞,1 ≤ kv1k∞,1kv2k∞,∞+kv1k∞,∞kv2k∞,1
≤C(kv1k1,∞,1kv2k∞,1+kv1k∞,1kv2k1,∞,1).
(5.2.2)
We further have
kwik∞,∞≤Ck∂zwik∞,1 ≤Ck∇Hvik∞,1 ≤Ckvik1,∞,1
and therefore
k∂z(w1v2)k∞,1 ≤ k∂zw1k∞,1kv2k∞,∞+kw1k∞,∞k∂zv2k∞,1
≤Ckv1k1,∞,1kv2k1,∞,1, kw1v2k∞,1 ≤ kw1k∞,∞kv2k∞,1
≤ kv1k1,∞,1kv2k∞,1.
(5.2.3)
Recall from (4.2.3) thatPf =f+ (R⊗R)f whereR⊗R= (RiRj)1≤i,j≤2 and f denotes the vertical average. By rewriting the bilinear term as
∇ ·(u1⊗v2) =∇H ·(v1⊗v2) +∂z(w1v2) and using the fact that wi vanishes for z = 0 andz =−h, we obtain
P∇ ·(u1⊗v2) =P∇H ·(v1⊗v2) +∂z(w1v2) + (RiRj)1≤i,j≤2∂z(w1v2)
=P∇H ·(v1⊗v2) +∂z(w1v2).
The case α = 0 then follows from Theorem 4.2.1 (v) and (vi), as well as (5.2.2) and (5.2.3). For the case α ∈ (0,1) we observe that the horizontal operators (−∆H)(1−α)/2, (−∆H)−(1−α)/2 and ∇H commute and therefore it holds that
S(t)P∇H ·(v1⊗v2) = S(t)P(−∆H)(1−α)/2∇H ·(−∆H)−(1−α)/2(v1⊗v2).
By Theorem 4.2.1.(iv), Lemma 3.2.3, and the estimates (5.2.2), we thus obtain kS(t)P∇H ·(v1⊗v2)k∞,1 ≤Ct−(1−α)/2k∇H ·(−∆H)−(1−α)/2(v1⊗v2)k∞,1
≤Ct−(1−α)/2kv1⊗v2k1−α∞,1k∇H(v1⊗v2)kα∞,1
≤Ct−(1−α)/2Gα(v1, v2).
For the remaining term, we observe that the fact that w1 vanishes forz =−h,0 implies that
S(t)∂z(w1v2) = S(t)∂zI1∂z(w1v2) =S(t)∂zIαI1−α∂z(w1v2) = S(t)∂zIα∂zα(w1v2), compare Definition 3.3.5, and so, by Theorem 4.2.1.(vi) and the estimates (5.2.3), we thus have
kS(t)∂z(w1v2)k∞,1 ≤Ct−(1−α)/2kI1−α(w1v2)k∞,1
≤Ct−(1−α)/2kw1v2k1−α∞,1k∂z(w1v2)kα∞,1
≤Ct−(1−α)/2Gα(v1, v2),
where in the last step we used that the interpolation inequality (ii) in Lemma 3.3.6 is also valid in L∞HLpz(L) by an obvious modification. This completes the proof.
Before we proceed to apply these estimates to the primitive equations, we also re-quire the following lemma concerning the boundedness of sequences subject to suitable recursive inequalities.
Lemma 5.2.6. Let c0, c1, c2 >0 be constant coefficients and let γ, δ >0 be such that 18c0c22γ ≤1 and 6c0δ ≤1.
Then there existsε >0such that any two sequences (αn)n∈N and (βn)n∈N of positive real numbers satisfying α0 ≤c0, β0 ≤ε, as well as the recursive growth bounds
αn+1 ≤c0+c1αnβn, βn+1 ≤ε+c2α1/2n βn3/2+δαnβn, are bounded with αn≤2c0 and βn≤γ for all n∈N.
Proof. We prove the estimates in two steps via induction. Let ε > 0 be so small that pε(x) := ε +c2(2c0)1/2x3/2 + x/3 has fixed points x0 = x0(ε) and x1 = x1(ε) with 0 < x0 < x1. Since limε→0x0(ε) = 0, we further take ε > 0 to be so small that x0 <1/2c1. Then we have
b0 ≤ε≤pε(x0) =x0 <1/2c1
and if it holds that αn ≤ 2c0 and βn ≤ x0 for some n ∈ N, then the recursive growth bounds imply that
αn+1 ≤c0+c12c0 1
2c1 ≤2c0, βn+1 ≤pε(βn)≤pε(x0) =x0 <1/2c1,
so by induction it follows that αn ≤ 2c0 and βn ≤ x0 <1/2c1 for all n ∈ N. Now take ε >0 to be so small that 3ε≤γ. Then we have β0 ≤γ and the recursive growth bound for (βn)n∈N yields that if it holds that βn≤γ for some n ∈N, then we also have
βn+1 ≤ε+c2√
2c0γ3/2+δ2c0γ
≤ 1
3+c2
p2c0γ+1 3
γ
≤γ, which implies the estimate for (βn)n∈N.
5.2 The case of Neumann boundary conditions Furthermore, we utilize the following Lemma concerning the continuity of convolution integrals.
Lemma 5.2.7. Let v ∈Cb((0,∞);Xσ∞,1(L))be a function such that t1/2∇v ∈Cb((0, T);L∞HL1z(L))2
and a ∈L∞,pσ (L). Then the function given by V(t) :=S(t)a−
Z t 0
S(t−s)P∇ ·(u⊗v)(s)ds satisfies V,∇V ∈C((0, T);L∞HL1z(L))2.
Proof. Let t, δ >0. Then we have for the first term V1(t) := S(t)a S(t+δ)a =S(δ)S(t)a,
∇HS(t+δ)a =S(δ)∇HS(t)a,
∂zS(t+δ)a =SH(δ)SD(δ)∂zS(t)a.
Here we used the fact that ∇H commutes with S(t) = SH(t)SN(t), that ∂z com-mutes with SH(t), as well as the elementary relation ∂zSN(t) = SD(t)∂z with SD from Lemma 3.3.2. SinceS acts on the horizontal component via convolution with the Gaus-sian kernel we have S(t)a,∇S(t)a∈BU C(R2;L1(−h,0))2 for t >0 and thus the conti-nuity V1 and ∇V1 follows from Theorem 4.2.1 and Lemma 3.3.2.
For the convolution term V2(t) := Rt
0 S(t−s)P∇ ·(u(s)⊗v(s))ds we have V2(t+δ)−V2(t) =I1(t, δ) +I2(t, δ),
with auxiliary terms
I1(t, δ) :=
Z t+δ t
S(t+δ−s)P∇ ·(u⊗v)(s)ds, and
I2(t, δ) :=
Z t 0
S(s)P∇ ·[(u⊗v)(t+δ−s)−(u⊗v)(t−s)] ds.
For I1 we have by Lemma 5.2.5 with α= 0 that kI1(t, δ)k∞,1 ≤C
Z t+δ t
(t+δ−s)−1/2kv(s)k1,∞,1kv(s)k∞,1ds.
Setting
K(t) := sup
0<s<t
s1/2kv(s)k1,∞,1, H(t) := sup
0<s<t
kv(s)k∞,1, we then have
kI1(t, δ)k∞,1 ≤CK(t+δ)H(t+δ) Z t+δ
t
(t+δ−s)−1/2s−1/2ds.
Replacing δ >0 with tδ >0, the latter integral can be further estimated via Z t+t·δ
t
(t+δ−s)−1/2s−1/2ds= Z 1+δ
1
(1 +δ−u)−1/2u−1/2du
≤ Z 1+δ
1
(1 +δ−u)−1/2du
= Z δ
0
x−1/2dx
= 2δ1/2,
where we used the substitutionsu=t·s andx= 1 +δ−u. This yieldsI1(t, tδ)→0 for δ→0.
For∇I1 we write∇S(t) =∇S(t/2)S(t/2) and apply Theorem 4.2.1.(i) and Lemma 5.2.5 with α= 1/2 to obtain
kI2(t, tδ)k∞,1 ≤CK(t+tδ)3/2H(t+tδ)1/2 Z t+tδ
t
(t+tδ−s)−3/4s−3/4ds.
Using the same substitutions yields Z t+tδ
t
(t+tδ−s)−3/4s−3/4ds=t−1/2 Z 1+δ
1
(1 +δ−u)−3/4u−3/4du
≤t−1/2 Z 1+δ
1
(1 +δ−u)−3/4du
=t−1/2 Z δ
0
x−3/4dx
=t−1/24δ1/4, and therefore ∇I1(t, tδ)→0 forδ →0 and t >0.
For the termI2(t, δ) we likewise apply Lemma 5.2.5 with α= 0, yielding kI2(t, δ)k∞,1 ≤C
Z t 0
s−1/2k(u⊗v)(t+δ−s)−(u⊗v)(t−s)k∞,1 ds.
Here we further have
(u⊗v)(t+δ−s)−(u⊗v)(t−s) =[u(t+δ−s)−u(t−s)]⊗v(t+δ−s) +u(t−s)⊗[v(t+δ−s)−v(t−s)], yielding
k(u⊗v)(t+δ−s)−(u⊗v)(t−s)k∞,1
≤Ckv(t+δ−s)−v(t−s)k1,∞,1kv(t+δ−s)k∞,1, +Ckv(t−s)k1,∞,1kv(t+δ−s)−v(t−s)k∞,1,
5.2 The case of Neumann boundary conditions
compare the proof of Lemma 5.2.5. We now utilize the estimate Z t
0
s−1/2kv(t+δ−s)−v(t−s)k1,∞,1kv(t+δ−s)k∞,1ds
≤K(t+δ) Z t
0
s−1/2(t−s)−1/2(t−s)1/2kv(t+δ−s)−v(t−s)k1,∞,1ds.
By our assumption on v, the term (t−s)1/2kv(t+δ−s)−v(t−s)k1,∞,1 vanishes for δ → 0 for almost alls ∈ (0, t) and is uniformly bounded for s ∈(0, t), δ ∈(0,1). Since the value of the integral
Z t 0
s−1/2(t−s)−1/2 = Z 1
0
u−1/2(1−u)−1/2du
is finite, it follows that the right-hand side vanishes forδ →0 by the dominated conver-gence theorem. The remaining term
Z t 0
s−1/2kv(t−s)k1,∞,1kv(t+δ−s)−v(t−s)k∞,1ds
is treated analogously. This yields∇I2(t, δ)→0 forδ →0. The remaining term∇I2(t, δ) is treated similarly via the same arguments we used for ∇I1(t, δ). This concludes the proof.
We are now able to prove our first main result.
Proof of Theorem 5.2.1. We construct the solutionvas the limit of the sequence (vn)n∈N, recursively defined by v0(t) := S(t)a and
vn+1(t) := S(t)a− Z t
0
S(t−s)P∇ ·(un⊗vn)(s)ds, n ∈N. For this purpose we consider the space
S(T) := {v ∈Cb((0, T);Xσ∞,1(L)) :t1/2∇v ∈L∞(0, T;L∞HL1z(L))2} endowed with the norm
kvkS(T):= max
sup
0<t<T
kv(t)k∞,1, sup
0<t<T
t1/2k∇v(t)k∞,1
.
The fact that this sequence belongs toS(T) for suitableT >0 follows from Lemma 5.2.7 together with the uniform estimates established in the next step.
Step 1: Uniform boundedness. We begin by establishing recursive inequalities for the quantities
Hn(t) := sup
0<s<t
kvn(s)k∞,1, Kn(t) := sup
0<s<t
s1/2kvn(s)k1,∞,1.
By Theorem 4.2.1 it holds for allt >0 that sup
0<s<t
kv0(s)k∞,1 ≤ kak∞,1, sup
0<s<t
s1/2k∇v0(s)k∞,1 ≤Ckak∞,1. Letn ∈Nand t >0 be arbitrary. We apply Lemma 5.2.5 forα= 0, yielding
kvn+1(t)k∞,1 ≤ kS(t)ak∞,1+C Z t
0
(t−s)−1/2kvn(s)k1,∞,1kvn(s)k∞,1ds
≤ kak∞,1+C1Hn(t)Kn(t)
(5.2.4)
for a constant C1 > 0, where we used that S is contractive and that the value of the integral
Z t 0
(t−s)−1/2s−1/2ds= Z 1
0
(1−s)−1/2s−1/2ds <∞ does not depend on t >0. This yields the estimate
Hn+1(t)≤ kak∞,1+C1Hn(t)Kn(t), (5.2.5) and by multiplying both sides in the first line of (5.2.4) we obtain
sup
0<s<t
s1/2kvn+1(s)k∞,1 ≤ sup
0<s<t
s1/2kS(s)ak∞,1+C1t1/2Hn(t)Kn(t). (5.2.6) For the gradient estimate, we split the derivative of the semigroup into
∇S(t−s) = ∇S
t−s 2
S
t−s 2
,
and so using the estimates (i) in Theorem 4.2.1 as well as Lemma 5.2.5 for α = 1/2 yields
k∇vn+1(t)k∞,1 ≤ k∇S(t)ak∞,1+C Z t
0
(t−s)−3/4kvn(s)k3/21,∞,1kvn(s)k1/2∞,1ds.
By multiplying both sides of this inequality by t1/2, it follows that sup
0<s<t
s1/2k∇vn+1(s)k∞,1 ≤ sup
0<s<t
s1/2k∇S(s)ak∞,1+C2Hn(t)1/2Kn(t)3/2 (5.2.7) since the value of the integral
t1/2 Z t
0
(t−s)−3/4s−3/4ds = Z 1
0
(1−s)−3/4s−3/4ds <∞
likewise does not depend on t > 0. We now add estimates (5.2.6) and (5.2.7) together to obtain
Kn+1(t)≤K0(t) +C2Hn(t)1/2Kn(t)3/2+C3t1/2Hn(t)Kn(t). (5.2.8)
5.2 The case of Neumann boundary conditions We now apply Lemma 5.2.6 to the sequences (Hn(t))n∈N and (Kn(t))n∈N. For this pur-pose, we note that
H0(t) = sup
0<s<t
kS(s)ak∞,1 ≤ kak∞,1
since S is contractive, as well as K0(t) = sup
0<s<t
s1/2kS(t)ak1,∞,1
≤t1/2kak∞,1+ sup
0<s<t
s1/2k∇S(t)a1k∞,1 + sup
0<s<t
s1/2k∇S(t)a2k∞,1
≤t1/2kak∞,1+ sup
0<s<t
s1/2k∇S(t)a1k∞,1 +C4ka2k∞,1,
where C4 > 0 is from estimate (i) in Theorem 4.2.1. By Theorem 4.2.1.4 and the continuity of a1, the first and second right-hand side terms converge to 0 for t → 0.
Therefore, we may take T0 =T0(kak∞,1, a1)>0 to be so small that T01/2kak∞,1+ sup
0<s<T0
s1/2k∇S(t)a1k∞,1 ≤C4ka2k∞,1
and this implies that K0(t)≤2C4ka2k∞,1 for all t∈(0, T0). We now choose the param-eters
c0 :=kak∞,1, c1 :=C1, c2 :=C2. Due to the assumption
kak∞,1· ka2k∞,1 ≤C0,
we may set γ := 3C4ka2k∞,1 and take C0, δ=δ(kak∞,1)>0 to be so small that 18c0c22γ = 54C22kak∞,1· ka2k∞,1 ≤54C0C22 ≤1, 6c0δ = 6δkak∞,1 ≤1.
We further take C0, T0 >0 to be so small that
K0(t)≤2C4ka2k∞,1 ≤2C4C0 ≤ε=ε(c0, c1, c2, γ, δ),
as well as C3T01/2 ≤δ. Then, by the recursive estimates (5.2.5) and (5.2.8), the assump-tions of Lemma 5.2.6 are satisfied and we obtain
Hn(t)≤2kak∞,1, Kn(t)≤3C4ka2k∞,1 (5.2.9) for all t∈(0, T0) andn ∈N. In particular, we have that (vn)n∈N is a bounded sequence in S(T).
Step 2: Convergence. Consider the auxiliary sequence (Vn)n∈N defined by Vn:=vn+1−vn, n ∈N,
as well as
Hn(t) := sup
0<s<t
kVn(s)k∞,1, Kn(t) := sup
0<s<t
s1/2kVn(s)k1,∞,1.
We use the representation Vn+1(t) =
Z t 0
S(t−s)P∇ ·(Un(s)⊗vn(s) +un(s)⊗Vn(s)) ds,
where Un = (Vn,Wn) and Wn is determined by Vn via the relation (5.0.2), and apply Lemma 5.2.5 forα = 1/2 to obtain
kVn+1(t)k∞,1 ≤2C Z t
0
(t−s)−1/4G1/2(vn(s),Vn(s))ds.
Since we have
G1/2(vn(s),Vn(s))≤s−3/4(Kn(t)Hn(t) +Hn(t)Kn(t))1/2Kn(t)1/2Kn(t)1/2 for all 0< s < t < T and the value of the integral
Z t 0
(t−s)−1/4s−3/4ds= Z 1
0
(1−s)−1/4s−3/4ds <∞ does not depend on t >0, it follows that
Hn+1(t)≤C5(Kn(t)Hn(t) +Hn(t)Kn(t))1/2Kn(t)1/2Kn(t)1/2. Proceeding analogously fort1/2kVn(t)k∞,1 and t1/2k∇Vn(t)k∞,1 then yields
Kn+1(t)≤C5(1 +t1/2) (Kn(t)Hn(t) +Hn(t)Kn(t))1/2Kn(t)1/2Kn(t)1/2. SettingNn(t) := max{Hn(t),Kn(t)}, we obtain the recursive estimate
Nn+1(t)≤C5(1 +T01/2) (Hn(t) +Kn(t))1/2Kn(t)1/2Nn(t) for all t∈(0, T0). Due to (5.2.9), we may take C0, T0 >0 to be so small that
Nn+1(t)≤ 1 2Nn(t)
for allt∈(0, T0) andn∈N. This implies that (vn)n∈Nis a Cauchy sequence inS(T0). A similar approach shows that (vn−Sa2)n∈N is a Cauchy sequence in C([0, T0];Xσ∞,1(L)).
Denoting the limit of (vn)n∈N by v, it is clear that v is a mild solution to the primitive equations, i.e., that it satisfies the integral equation (5.0.6). The gradient estimate
lim sup
t→0
t1/2k∇v(t)k∞,1 ≤Cka2k∞,1
follows from (5.2.9).
5.2 The case of Neumann boundary conditions Step 3: Uniqueness. Let ˜v be another such solution on [0, T0] and consider the auxiliary quantities
H(t) := sup0<s<tkv(s)−˜v(s)k∞,1, K(t) := sup0<s<ts1/2k∇v(s)− ∇˜v(s)k∞,1, H(t) := sup0<s<tkv(s)k∞,1, K(t) := sup0<s<ts1/2k∇v(s)k∞,1,
H(t) := sup˜ 0<s<tk˜v(s)k∞,1, K(t) := sup˜ 0<s<ts1/2k∇˜v(s)k∞,1. Then the same arguments as in the previous step yield the estimates
H(t)≤C5(H(t)K(t) +K(t)H(t))1/2K(t)1/2K(t)1/2 +C5( ˜H(t)K(t) + ˜K(t)H(t))1/2K(t)˜ 1/2K(t)1/2 and
K(t)≤C5(1 +t1/2)(H(t)K(t) +K(t)H(t))1/2K(t)1/2K(t)1/2 +C5(1 +t1/2)( ˜H(t)K(t) + ˜K(t)H(t))1/2K(t)˜ 1/2K(t)1/2. Setting N(t) := max{H(t),K(t)}, this yields
N(t)≤C5(1 +T01/2)
(H(t) +K(t))1/2K(t)1/2+ ( ˜H(t) + ˜K(t))1/2K(t)˜ 1/2
N(t).
(5.2.10) By our assumption on the solutions v and ˜v we have
limt→0K(t)≤C6ka2k∞,1, lim
t→0
K(t)˜ ≤C6ka2k∞,1. The same argument used to derive (5.2.5) then yields
limt→0
H(t)˜ ≤ kak∞,1+C1H(t) ˜˜ K(t) and thus
H(t)≤2kak∞,1, H(t)˜ ≤2kak∞,1
for all t ∈ (0, T1) if T1 > 0 is sufficiently small, where we also used (5.2.9). Applying these estimates to (5.2.10) yields
N(t)≤C6(1 +T11/2) (kak∞,1+ka2k∞,1)ka2k1/2∞,1N(t) (5.2.11) for t ∈(0, T1), so by taking T0 >0 and the upper bound C0 >0 to be sufficiently small we obtain N(t) ≤ 12N(t) for all t ∈ (0, T0] and this yields v(t) = ˜v(t) for all t ∈ [0, T0].
Now let
t∗ := sup{t∈(0, T0) :v(s) = ˜v(s) for all 0≤s ≤t}
and assume that t∗ < T0. By the argument above we have t∗ ≥T1 >0 and thus a∗ :=v(t∗) = ˜v(t∗)
by the continuity of solutions on (0, T0). We now repeat the argument for the new initial value a∗. Since t1/2∇v, t1/2∇˜v ∈L∞((0, T0);L∞HL1z(L))2 yields
∇v,∇˜v ∈L∞((t∗, T0);L∞HL1z(L))2, we obtain
lim sup
t→0
t1/2k∇v(t∗ +t)k∞,1 = lim sup
t→0
t1/2k∇˜v(t∗ +t)k∞,1 = 0 as well as
lim sup
t→0
kv(t∗+t)k∞,1 ≤2ka∗k∞,1, lim sup
t→0
k˜v(t∗+t)k∞,1 ≤2ka∗k∞,1.
It follows that estimate (5.2.11) also holds for all t∈(t∗, T2) witht∗ < T2 < T0. Taking C0 > 0 to be sufficiently small we thus obtain v = ˜v on (0, T2) which contradicts the definition of t∗ and thus we have v = ˜v on (0, T0). This completes the proof.
We now prove the additional regularity of the solution for initial data belonging to L∞HLpz(L)2.
Proof of Proposition 5.2.4. We write S(t) = S(t/2)S(t/2) and use the vertical L1-Lp -smoothing estimate forS from Theorem 4.2.1.(b), the fact that S is contractive as well as Lemma 5.2.5 for α= 0 to obtain
kv(t)k∞,p ≤ kak+ Z t
0
1 + (t−s)−(1−1/p)/2
(t−s)−1/2s−1/2ds
K(t)H(t) withK(t) := sup0<s<tkv(s)k1,∞,1 and H(t) := sup0<s<tkv(s)k∞,1. Since the values of the integrals
Z t 0
(t−s)−1/2s−1/2ds, t(1−1/p)/2 Z t
0
(t−s)−(1−1/p)/2(t−s)−1/2s−1/2ds
do not depend ont >0, the estimate fort(1−1/p)/2v follows. For the gradient estimate we write ∇S(t) = S(t/3)S(t/3)S(t/3) and use Theorem 4.2.1.(i), the L∞HL1z(L)-L∞HLpz(L )-smoothing estimate and Lemma 5.2.5 forα∈(1−1/p,1) to obtain
k∇v(t)k∞,p ≤ k∇S(t)ak∞,p
+C Z t
0
1 + (t−s)−(1−1/p)/2
(t−s)−(1−α/2)s−(1+α)/2dsK(t)1+αH(t)1−α. Since the values of the integrals
t1/2 Z t
0
(t−s)−(1−α/2)s−(1+α)/2ds, t1−1/2p Z t
0
(t−s)−(1−1/p)/2−(1−α/2)
s−(1+α)/2ds likewise do not depend on t > 0 and are finite for α ∈ (1−1/p,1), the estimate for t1−1/2p∇v follows as well. In the cases (ii) and (iii) the continuity at t= 0 follows from the strong continuity ofS, and in the case (iii) the continuity in the variables (x, y, z)∈L follows from the fact that bothSH and SN preserve continuity.
5.2 The case of Neumann boundary conditions With Proposition 5.2.4 and our result for local existence, we are now able to derive our second main result concerning global, strong well-posedness.
Proof of Theorem 5.2.2. Letabe horizontally periodic. SinceSH andPand the nonlin-ear term (u· ∇)v preserve horizontal periodicity, the way we constructed the solution v in the proof of Theorem 5.2.1 yields that v is also horizontally periodic. We may assume without loss of generality that the period is 1. Since we have v(t),∇v(t) ∈ L∞HL1z(L)2 for all t >0, the embeddings
W1,1(−h,0)⊂Lp(−h,0) and L∞(G),→Lp(G) for G= (0,1)2 and allp∈(1,∞) imply that
v(t)
Ω ∈H1,p(Ω)2
for all t > 0 and Ω = G ×(−h,0). Due to the global, strong well-posedness of the primitive equations with Neumann boundary conditions on Γu∪Γb as in (3.4.2), we take v(t0) for t0 >0 as the initial data and obtain a strong extension onto Ω×(t0,∞) that is even real analytic by Theorem 5.1.4. Due to the uniqueness of mild solutions and horizontal periodicity, we obtain global, strong well-posedness on L×(0,∞).